Question

If the quadratic formula is used to solve 2x^2 - 3x - 1 = 0, what are the solutions?

Answer 1

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{-3}x\stackrel{\stackrel{c}{\downarrow }}{-1}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x= \cfrac{ - (-3) \pm \sqrt { (-3)^2 -4(2)(-1)}}{2(2)} \implies x = \cfrac{ 3 \pm \sqrt { 9 +8}}{ 4 } \\\\\\ x= \cfrac{ 3 \pm \sqrt { 17 }}{ 4 }\implies x= \begin{cases} \cfrac{ 3 + \sqrt { 17 }}{ 4 }\\\\ \cfrac{ 3 - \sqrt { 17 }}{ 4 } \end{cases}`

Answer 2

`\displaystyle x=(3+√(17) )/(4) ,(3-√(17) )/(4)`

Explanation:

The quadratic formula is a method of solving for x in a quadratic function.

Quadratic Formula

Remember that quadratic functions are written as ax^2 + bx + c. The quadratic formula is
`(-b\pm√(b^2-4ac) )/(2a)` and is used when the function is equal to 0. So, for this function: a = 2, b = -3, and c = -1.

Solving For x

That means the quadratic formula will look like
`\displaystyle (3\pm√(9-(4*2*-1)))/(2*2)`. To find x, first, we can simplify the denominator.

• 2*2=4

Next, we can first solve the expression under the square root.

• 
  `√(9-(4*2*-1))`=
  `√(9+8)`
• 
  `√(17)`

Then, since this is not a perfect square, we can just put this back into the equation to get the final answer.

• 
  `\displaystyle (3\pm√(17))/(4)`